2x^2+10x= eching

$$2 x ^ { 2 } + 10 x =$$

Omil

$$2x\left(x+5\right)$$

Yechim qadamlari

2 omili.

$$2\left(x^{2}+5x\right)$$

Hisoblang: x^{2}+5x. x omili.

$$x\left(x+5\right)$$

Toʻliq ajratilgan ifodani qaytadan yozing.

$$2x\left(x+5\right)$$

Kvadrat tenglamalaridan foydalanish qoidalari

Kvadrat koʻp tenglama bu orqali hisoblanadi: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), bu yerda x_{1} va x_{2} ax^{2}+bx+c=0 kvadrat tenglamaning yechimlari.

$$2x^{2}+10x=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$x=\frac{-10±\sqrt{10^{2}}}{2\times 2}$$

10^{2} ning kvadrat ildizini chiqarish.

$$x=\frac{-10±10}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-10±10}{4}$$

x=\frac{-10±10}{4} tenglamasini yeching, bunda ± musbat. -10 ni 10 ga qo'shish.

$$x=\frac{0}{4}$$

0 ni 4 ga bo'lish.

$$x=0$$

x=\frac{-10±10}{4} tenglamasini yeching, bunda ± manfiy. -10 dan 10 ni ayirish.

$$x=-\frac{20}{4}$$

-20 ni 4 ga bo'lish.

$$x=-5$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 0 ga va x_{2} uchun -5 ga bo‘ling.

$$2x^{2}+10x=2x\left(x-\left(-5\right)\right)$$

p-\left(-q\right) shaklining barcha amallarigani p+q ga soddalashtiring.

$$2x^{2}+10x=2x\left(x+5\right)$$

Baholash

$$2x\left(x+5\right)$$